same here, If there were a poll, my guess is that an overwhelming majority of users would say that.

Maple works very well, the problem is we don't seem to understand your question. If you did something like this in Mathematica, why not post the Mathematica code here, maybe then your question will become clearer.

Maple works very well, the problem is we don't seem to understand your question. If you did something like this in Mathematica, why not post the Mathematica code here, maybe then your question will become clearer.

thanks Robert, I couldn't have gone this far without your help in previous mapleprimes exchanges!

The point at which I'm stuck at present is that of turning your transformed system "newsys" into the so-called "canonical" or "normal" form, which consists in using the eigenvectors somehow.  (I can do simple 2D cases by hand, but I get stuck for this 3D system even with Maple's help)...

thanks Robert, I couldn't have gone this far without your help in previous mapleprimes exchanges!

The point at which I'm stuck at present is that of turning your transformed system "newsys" into the so-called "canonical" or "normal" form, which consists in using the eigenvectors somehow.  (I can do simple 2D cases by hand, but I get stuck for this 3D system even with Maple's help)...

not enough if your purpose is to fit a curve through data points. See below two possibilities (among many) based on the information you have given.

restart:
with(plots):

pointplot ([ [-60,-1], [-50,-0.9], [-40,-0.8], [-30,-0.7], [-20,-0.6], [-10,-0.5], [0,-0.4], [10,-0.3], [20,-0.2], [30,-0.1], [40,0], [50,0.1], [60,0.2]]);

pointplot ([ [-60,1], [-50,0.9], [-40,0.8], [-30,0.7], [-20,0.6], [-10,0.5], [0,0.4], [10,0.3], [20,0.2], [30,0.1], [40,0], [50,-0.1], [60,-0.2]]);

Now perhaps I do not understand your question to begin with.

not enough if your purpose is to fit a curve through data points. See below two possibilities (among many) based on the information you have given.

restart:
with(plots):

pointplot ([ [-60,-1], [-50,-0.9], [-40,-0.8], [-30,-0.7], [-20,-0.6], [-10,-0.5], [0,-0.4], [10,-0.3], [20,-0.2], [30,-0.1], [40,0], [50,0.1], [60,0.2]]);

pointplot ([ [-60,1], [-50,0.9], [-40,0.8], [-30,0.7], [-20,0.6], [-10,0.5], [0,0.4], [10,0.3], [20,0.2], [30,0.1], [40,0], [50,-0.1], [60,-0.2]]);

Now perhaps I do not understand your question to begin with.

plot(cos(x)-x, x = - 10^(-16)... 10^(-16));

 Maple 13.01

plot(cos(x)-x, x = - 10^(-16)... 10^(-16));

 Maple 13.01

what data?

what data?

that's right, you have too many "degrees of freedom", in other words not enough constraint on the problem to meaninfully start a search for A and B...

that's right, you have too many "degrees of freedom", in other words not enough constraint on the problem to meaninfully start a search for A and B...

somehow just came across this two-year old thread. Allow me to revive it.

I read the French article quoted above on the etymology of the word "martingale" - it's very entertaining.

Let me summarize it briefly: there is no agreement about the etymology of the word; it's most likely derived from the name of the city of Martigues in the south of France (near Marseille); the earliest traces of its usage in French,  in the context of gambling, can be dated to the middle of the 18th century; Casanova, in his memoirs (if you haven't read them, read them now), used the strategy and named it martingale.

Some highlights, at random:

Definition from the French Academy, 1762, one of the earliest definitions in the French language:

"Jouer à la martingale, c'est jouer toujours tout ce que l'on a perdu" --> to play a martingale [strategy] consists in always gambling [amounts equivalent to] your losses.

The most convincing of the stories is the one that traces the expression "to play a martingale" to the Provençal language (a nearly extinct dialect, formerly language):

Une piste, a priori ténue, semble faire dériver ce mot de l'expression provençale "jouga a la martegalo" qui signifierait "jouer de manière incompréhensible, absurde" --> a believable theory is that the word derives from the Provençal expression "jouga a la martegalo" which would mean "to play in an absurd, incomprehensible manner". That is, if we interpret a little, the strategy of "doubling your bets" was thought of as crazy. Then it goes on to say that the people from Martigues were thought of as "naive" (read "stupid") and so, the story goes, to "play a martingale" is "to play like an inhabitant of Martigues, notorious for being silly".

There is another candidate etymology, also related to Martigues. A special type of clothing had that name, and it consisted in a cloth opened at the back and tied by a leather belt, hence the name martingale also associated with leather belts, and hence the association with equipment for horse-riding. This theory would be boring were it not for a quotation of Rabelais, a 16th century French writer who liked to curse. Rabelais describes the piece of clothing thus:

"un pont-levis de cul pour plus aisément fienter" -> "a drawbridge for the arse to defecate comfortably"

We have now drifted a little from probability strategies, but as someone who grew up in Provence, I couldn't resist sharing this stuff with non-French speakers.

I understood your suggested change of variable to be the following, but perhaps I misunderstood -- I now realize your system is a little different (some square root on x has disappeared), so I'm not sure if the following is what you meant:

newxdot := diff(x(tau),tau) = rhs(eval(xdot,t=tau))*rhs(eval(xdot,t=tau)):

newcdot := diff(c(tau),tau) = rhs(eval(cdot,t=tau))*rhs(eval(xdot,t=tau)):

newqdot := diff(q(tau),tau) = rhs(eval(qdot,t=tau))*rhs(eval(xdot,t=tau)):

If it's not what you meant, what I wrote probably doesn't make sense.

Remark. Hopefully mapleprimes will respect my linebreaks this time.